All performance data for a/c is given in relation to the standard atmosphere. We must be able to convert ambient conditions prevailing at a particular aerodrome to an equivalent height in ISA.
Pressure Ht= Ht in the standard atmosphere where the pressure is the same as the sample of air being considered. (It is a measure of PRESSURE, not Height!)
At a lower QNH the field elevation must be higher than Sea Level as the air density is lower.
Example:
Field elevation of an aerodrome is 1500 Ft. Atm pressure at S.L. on that day is 1000 hPa. At what level in the std atm would the air be the same pressure as the air in the field?
We are asked to find the PRESSURE HEIGHT of the field.
Want to know the height of the field, not above the sea (on that day the QNH is 1000 hPa) , but above the 1013 hPa pressure level which is the starting point for the std atm.
1013 is greater than the QNH of the day, 1000 hPa; therefore it must be further down than S.L. since pressure in the atm can only increase with decreased ht (1013 is > 1000).
1013 must be (1013-1000)=13 x 30 Ft = 390 Ft below the sea on that day.
Hence the ht of the field above the 1013 pressure level is 1500+390=1890 Ft. This is the Press Ht. of the field. The airplane will behave and perform according to the 1890 Ft figures and not the 1500 figures.
Pressure Ht= Ht in the standard atmosphere where the pressure is the same as the sample of air being considered. (It is a measure of PRESSURE, not Height!)
At a lower QNH the field elevation must be higher than Sea Level as the air density is lower.
Example:
Field elevation of an aerodrome is 1500 Ft. Atm pressure at S.L. on that day is 1000 hPa. At what level in the std atm would the air be the same pressure as the air in the field?
We are asked to find the PRESSURE HEIGHT of the field.
Want to know the height of the field, not above the sea (on that day the QNH is 1000 hPa) , but above the 1013 hPa pressure level which is the starting point for the std atm.
1013 is greater than the QNH of the day, 1000 hPa; therefore it must be further down than S.L. since pressure in the atm can only increase with decreased ht (1013 is > 1000).
1013 must be (1013-1000)=13 x 30 Ft = 390 Ft below the sea on that day.
Hence the ht of the field above the 1013 pressure level is 1500+390=1890 Ft. This is the Press Ht. of the field. The airplane will behave and perform according to the 1890 Ft figures and not the 1500 figures.
PRESSURE HT= ELEVATION + (1013-QNH) X 30
PRESSURE HT INDICATES THE HEIGHT IN THE STANDARD ATMOSPHERE WHERE THE PRESSURE IS THE SAME AS THAT AT THE LOCATION BEING CONSIDERED.
PRESSURE HT INDICATES THE HEIGHT IN THE STANDARD ATMOSPHERE WHERE THE PRESSURE IS THE SAME AS THAT AT THE LOCATION BEING CONSIDERED.
Density Height
is the height in the std atm (ISA) where the air has the same density as the sample of air being considered. (It is a measure of density, not height!)
Press Ht concerns with air density changes due to atm pressure. We need to consider air density changes due to temp also.
ISA Temp - What is the behavior of temp in the std atm? SL std temp is 15 C. As temp drops at a std rate of 2 C per 1000' of height increase we can then calculate the std temp in ISA at any level.
E.g. at 4,000 Ft in the std atm the temp must be 15 - 2 x 4 = 15 - 8 = 7 C
This is called ISA Temp.
Temp Deviation - As air temp increases, air density decreases. Since density also decreases with increased ht, increases in air temp have a similar effect on density as an increase in ht ie an a/c will perform on a HOT day as though it is operated at a higher level in the std atm. Conversely it will operate on a cold day as though it is at a lower level.
Each deg C of temp rise = density decrease equiv to a ht increase of 120 Ft
E.g. on a hot day when the temp is 10 C hotter than ISA the a/c feels like it is operating at 10 x 120' = 1200 Ft higher.
Each deg C of temp drop = density increase equiv to a ht decrease of 120 Ft.
To calculate Density Ht we need pressure (i.e. Pressure Ht) and temp deviation from std atm:-
Example:
Find the density ht of an aerodrome which has an elevation of 2100 Ft if the QNH is 1000 hPa and the temp at the field at the time is +25C.
1) Given elevation and QNH we can find Pressure Height.
1013-1000=13
13 x 30 = 390 Ft
Press Ht is then 2100 + 390 = 2490 Ft
2) Temp drops at a std rate of 2 C per 1000 Ft rise.
Use Press Ht of approximately 2500 rounded off to nearest 500 Ft. So temp at a pressure ht of 2500 Ft must be colder than ISA:
ISA temp is 15 - (2 x 2.5) = +10C
Actual temp at the field is 25 C, therefore ISA Temp is COLDER than OAT hence PLUS.
ISA deviation is PLUS 25-10=+15 C
3) Multiply ISA deviation by 120 to obtain the correction that must be applied to the Pressure Ht to allow for the non std temp.
120 x ISA deviation = 1800 Ft
Density Height is then 2490 + 1800 = 4290 Ft//
is the height in the std atm (ISA) where the air has the same density as the sample of air being considered. (It is a measure of density, not height!)
Press Ht concerns with air density changes due to atm pressure. We need to consider air density changes due to temp also.
ISA Temp - What is the behavior of temp in the std atm? SL std temp is 15 C. As temp drops at a std rate of 2 C per 1000' of height increase we can then calculate the std temp in ISA at any level.
E.g. at 4,000 Ft in the std atm the temp must be 15 - 2 x 4 = 15 - 8 = 7 C
This is called ISA Temp.
Temp Deviation - As air temp increases, air density decreases. Since density also decreases with increased ht, increases in air temp have a similar effect on density as an increase in ht ie an a/c will perform on a HOT day as though it is operated at a higher level in the std atm. Conversely it will operate on a cold day as though it is at a lower level.
Each deg C of temp rise = density decrease equiv to a ht increase of 120 Ft
E.g. on a hot day when the temp is 10 C hotter than ISA the a/c feels like it is operating at 10 x 120' = 1200 Ft higher.
Each deg C of temp drop = density increase equiv to a ht decrease of 120 Ft.
To calculate Density Ht we need pressure (i.e. Pressure Ht) and temp deviation from std atm:-
Example:
Find the density ht of an aerodrome which has an elevation of 2100 Ft if the QNH is 1000 hPa and the temp at the field at the time is +25C.
1) Given elevation and QNH we can find Pressure Height.
1013-1000=13
13 x 30 = 390 Ft
Press Ht is then 2100 + 390 = 2490 Ft
2) Temp drops at a std rate of 2 C per 1000 Ft rise.
Use Press Ht of approximately 2500 rounded off to nearest 500 Ft. So temp at a pressure ht of 2500 Ft must be colder than ISA:
ISA temp is 15 - (2 x 2.5) = +10C
Actual temp at the field is 25 C, therefore ISA Temp is COLDER than OAT hence PLUS.
ISA deviation is PLUS 25-10=+15 C
3) Multiply ISA deviation by 120 to obtain the correction that must be applied to the Pressure Ht to allow for the non std temp.
120 x ISA deviation = 1800 Ft
Density Height is then 2490 + 1800 = 4290 Ft//
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